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Nature of Roots

If difference...

Question

If difference of roots of the equation $x^{2} - p x + q = 0$ is $1,$ then $p^{2} + 4 q^{2}$ equals-

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Solution

Consider the given equation.
$x^{2} - p x + q = 0$ $. . . . . . . . (1)$

So,
$α + β = p$
$α β = q$

Given,
$α - β = 1$

We know that
$(α + β)^{2} = (α - β)^{2} + 4 α β$

Therefore,
$p^{2} = 1^{2} + 4 q$
$p^{2} - 4 q = 1$

Hence, this is the answer.

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x^{2} - p x + q = 0

p^{2} - 4 q = 1

p^{2} + 4 q^{2} = {(1 + 2 q)}^{2}

x^{2} - p x + q = 0

p^{2} + 4 q = 1

p^{2} - 4 q = 1

p^{2} + 4 q^{2} = (1 + 2 q)^{2}

4 p^{2} + q^{2} = (1 + 2 p)^{2}

s e c α, c o s e c α

x^{2} - p x + q = 0

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State the following statement is True or False

If the roots of the equation $x^{2} + p x + q = 0$ differ by $1$ , then $p^{2} = 1 + 4 q$

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